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class Solution {
public:
    bool isSelfCrossing(vector<int>& x) {
        for(int i=3, l=x.size(); i<l; i++) {
            // Case 1: current line crosses the line 3 steps ahead of it
            if(x[i]>=x[i-2] && x[i-1]<=x[i-3]) return true;  
            // Case 2: current line crosses the line 4 steps ahead of it
            else if(i>=4 && x[i-1]==x[i-3] && x[i]+x[i-4]>=x[i-2]) return true; 
            // Case 3: current line crosses the line 6 steps ahead of it
            else if(i>=5 && x[i-2]>=x[i-4] && x[i]+x[i-4]>=x[i-2] && x[i-1]<=x[i-3] && x[i-1]+x[i-5]>=x[i-3]) return true;  
        }
        return false;
    }
 
/*               i-2
    case 1 : i-1┌─┐
                └─┼─>i
                 i-3
 
                    i-2
    case 2 : i-1 ┌────┐
                 └─══>┘i-3
                 i  i-4      (i overlapped i-4)
 
    case 3 :    i-4
               ┌──┐ 
               │i<┼─┐
            i-3│ i-5│i-1
               └────┘
                i-2
 
*/

/*
There are only 3 scenarios where it won’t cross itself.
The distances of the moves parallel to each other keeps going up (growing spiral).
The distances of the moves parallel to each other keeps going down (shrinking spiral).
The distances of the moves parallel to each other first keeps going up, then keeps going down (shrinking spiral inside of the growing spiral), and never goes up.
from  https://leetcode.com/discuss/88038/java-o-n-o-1-0ms-solution-with-explanation
不交叉的有以下三种情况
平行移动的距离是不断的增加的（螺旋形上升）
平行移动的距离是不断的减少的（螺旋形下降）
平行移动的距离先增加后减少，并且再也不增加。
*/

public class Solution {
    public boolean isSelfCrossing(int[] x) {
        int n = x.length;
		if (n < 4) return false;
		int t1 = 0, t2 = x[0], t3 = x[1], t4 = x[2], t5;
		boolean increase = t4 > t2 ? true : false;
		for (int i = 3; i < n; i++) {
			t5 = x[i];
			if (increase && t3 >= t5) {
				if (t5 + t1 < t3 || i + 1 < n && x[i + 1] + t2 < t4)
					increase = false;
				else if (i + 1 < n)
					return true;
			}
			else if (!increase && t3 <= t5)
				return true;
			t1 = t2;
			t2 = t3;
			t3 = t4;
			t4 = t5;
		}
		return false;
    }
}